Complete the assignment:
Objective:
You are to calculate a Z-score, determine the percentage in specification, and the percentage expected to be out of spec. The team has done some research that suggests that the factors they are studying will not greatly impact the standard deviation (spread) of student scores. However, they can significantly impact the mean score. The team must determine how far the mean or average student test score must shift in order to retain the federal funding.
Instructions :
Calculate the average student score necessary for the district to retain its federal funding. (You may assume the standard deviation will not change). This will require some thought. Think of the z formula. You will be solving for μ (average test score in the next school year.)
Think about what you already know. You know the standard deviation (because we assume that the variation in test scores stays approximately the same across years.)
You also know that 70% of students must meet the cut off score.
Data:
The following are the results from last year's MEAP scores for EHS
N = 1000
μ = 69.7
σ = 11.55
A student must score at or above 70% on the MEAP to meet the president's cut off score
Objective:
We want you to calculate a chi-square statistic and compare it with the critical value to determine whether or not GPA is significantly related to another of the factors being studied (so as not to completely discount the survey data). They chose to examine the relationship between the GPA of students and the teacher they had. They need to determine whether or not teacher choice impacts a student's GPA.
Instructions:
1. Utilize the data provided to determine with 95% confidence whether or not the choice of teacher affects a student's GPA.
2. Write up the null hypothesis statement. "Hypothesis statement?"
3. Calculate the chi-square statistic and compare that with the critical value for chi-square at a 95% confidence. "Write-up my conclusion?"
4. Write up your hypothesis conclusion.
Data:
# of Students w/ Given GPA
Teacher 2.01 - 2.50 2.51 - 3.00 3.01 - 3.50 3.51 - 4.00
A 68 74 73 78
B 70 73 75 77
C 69 75 72 76
You may want to review this practice example: You first need to calculate the expected values. It is best to make a table as step 1. For example, males and females watch various TV stations. Let's say that we want to find out if gender is dependent or independent of television station preferences. The practice data follows:
WKBW WBEN WGR Totals
Males 62 54 25 141
Females 44 50 15 109
Totals 106 104 40 250
Step 1. Calculate each of the 'expected values.' We will do
the first two for you.
a. Probability of viewer being male is 141 / 250 = 0.564 (refer to the cells in the table above)
b. Probability of viewer preferring WKBW is 106 / 250 = 0.424
c. Probability of viewer preferring WKBW AND being male is 0.564 x 0.424 = 0.239
d. Expected number of viewers in this cell is 0.239 x 250 = 59.8 etc...
a. Probability of viewer being female is 109 / 250 = 0.436
b. Probability of viewer preferring WKBW is 106 / 250 = 0.424
c. Probability of viewer preferring WKBW AND being female is 0.436 x 0.424 = 0.185
d. Expected number of viewers in this cell is 0.185 x 250 = 46.3 etc...
Repeat this for all six cells. To check your work, the totals (across and down) should add up very close to the (across and and down) totals of the observed values. Here is how your chart should appear when finished: The 'expected values are in [brackets.]
WKBW WBEN WGR Totals
Males 62 [59.8] 54 [58.8] 25 [22.5] 141
Females 44 [46.3] 50 [45.3] 15 [17.5] 109
Totals 106 104 40 250
Step 2. Compare the OBSERVED [EXPECTED]
Example: For the first cell (Males/WKBW), the formula is:
Observed minus [Expected] Squared divided by [Expected] as follows:
(62-59.8)2 divided by 59.8 = 0.0809
For the 2nd value... (54-58.8)2 divided by 58.8 = 0.3918
For the 3rd value... (25-22.5)2 divided by 22.5 = 0.2778
For the 4th value... (44-46.3)2 divided by 46.3 = 0.1143
For the 5th value... (50-45.3)2 divided by 45.3 = 0.4876
For the 6th value... (15-17.5)2 divided by 59.8 = 0.3571
Step 3. Add those chi-square values and you should get 1.7095 This is your calculated chi-square statistic.
Step 4. Determine the significance level. (e.g., .05 or .01 or .1) This is up to the discretion of the team and the team's choice is based
upon what level of risk they are willing to live with.
Step 5. Determine the Degrees of Freedom (df) for the rows and the columns. You will need this to find the critical value
df=(number of rows minus 1) multiplied by the (number of columns minus 1) So, in this particular case it would be (2-1 multiplied by 3-1) = 2
Step 6. Go to the chi-square table in a textbook and determine the critical value. You already know the calculated statistic (1.7095). All you need now is that critical value to see if the calculated statistic is greater than that critical value. If it is greater than the critical value, you can reject the null. If not, then you are forced to accept the null.
Objective:
The team determined that to confirm the validity of their screening experiment they should compare the test score results for some students that were at both the high setting and low setting within the last three years. This would allow them to compare the impact of the factors for individual students.
Instructions:
For each of the three factors for which data is provided:
a. Write a null and alternative hypothesis statement about the impact of the factor on student test scores.
b. Test each of the hypotheses at 90% confidence level. Note: you should know the tool to use. If you are struggling deciding which analysis tool to use contact your instructor for guidance. Which tool?
c. Based on the results of steps a. and b., determine whether or not each of three factors have a significant impact on student test scores. Include your statistic and critical values with your conclusion.
Data:
STYLE COMM INVOLV 2 PARENT HOME
Student Lecture Groups Student No Yes Student No Yes
1 67 71 1 67 75 1 68 74
2 75 77 2 73 81 2 71 78
3 54 52 3 58 63 3 54 57
4 81 79 4 81 84 4 85 81
5 72 76 5 69 64 5 71 75
6 77 83 6 82 84 6 80 82
7 89 91 7 65 69 7 62 59
8 59 67 8 78 74 8 77 77
9 63 72 9 80 85 9 83 90
10 71 75 10 71 74 10 68 71
11 82 81 11 81 83 11 80 78
12 64 67 12 67 72 12 60 64
13 74 74 13 71 66 13 74 69
14 84 89 14 82 79 14 83 86
15 69 78 15 74 79 15 78 75
Attachment:- Null hypothesis.rar